G r a d e 11 P r e -C a l C u l u s M a t h e M at i C s (30s)6 of 26 Grade 11 Pre-Calculus Mathematics 3. Write the defining linear function of the following arithmetic sequence.

G r a d e 1 1 P r e - C a l C u l u s M a t h e M a t i C s ( 3 0 s )

Midterm Practice Exam Answer Key

M i d t e r m P r a c t i c e E x a m A n s w e r K e y 3 of 26

G r a d e 1 1 P r e - C a l C u l u s M a t h e M a t i C s

Midterm Practice Exam Answer Key

Name: ___________________________________

Student Number: ___________________________

Attending q Non-Attending q

Phone Number: ____________________________

Address: _________________________________

__________________________________________

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Instructions

The midterm examination will be weighted as follows: Modules 1–4 100%The format of the examination will be as follows: Module 1: Sequences and Series 24 marks Module 2: Factoring and Rational Expressions 23 marks Module 3: Quadratic Functions 25 marks Module 4: Solving Rational and Quadratic Equations 28 marksTime allowed: 2.5 hours

Note: You are allowed to bring the following to the exam: pencils (2 or 3 of each), blank paper, a ruler, a scientific or graphing calculator, and your Midterm Exam Resource Sheet. Your Midterm Exam Resource Sheet must be handed in with the exam. You will receive your Midterm Exam Resource Sheet back from your tutor/marker with the next module work that is submitted for marking.

Show all calculations and formulas used. Include units where appropriate. Clearly state your final answer.

For Marker’s Use Only

Date: _______________________________

Final Mark: ________ /100 = ________ %

Comments:

Answer Key

G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s4 of 26


Name:

Answer all questions to the best of your ability. Show all your work.

Module 1: Sequences and Series (24 marks)

1. For each sequence below, indicate whether it is geometric, arithmetic, or neither. If it is geometric, state the value of r (the common ratio). If it is arithmetic, state the value of d (the common difference). (1 mark each, for a total of 3 marks) (Lesson 3)

a)

81 9 119

, , , , . . .

Answer:

Geometric sequence with a common ratio of

19

.

b) 1, 6, 9, 10, . . . Answer: Neither.

c) 1, 4, 7, 10, . . . Answer: Arithmetic sequence with a common difference of 3.

2. Explain the relationship between arithmetic sequences and linear functions. (2 marks) (Lesson 1)

Answer: If the domain of a linear function is the set of natural numbers (or positive integers), the

range is called an arithmetic sequence. Also, the common difference of an arithmetic sequence is the same as the slope in the corresponding linear function.


3. Write the defining linear function of the following arithmetic sequence. (2 marks) (Lesson 1) 3, 8, 13, . . .

Answer: One point on line: (1, 3) Slope: d = 8 – 3 = 5 Defining linear function: y – y1 = m(x – x1) y – 3 = 5(x – 1) y = 5x – 5 + 3 y = 5x – 2

4. Which term of the arithmetic sequence -4, 2, 8, . . . is the number 170? (3 marks) (Lesson 1)

Answer: Method 1: Using the general term formula tn = t1 + (n – 1)d: 170 = –4 + (n – 1)(6) 170 + 4 = 6n – 6 174 + 6 = 6n 180 = 6n 30 = n Therefore, the 30th term is 170.

Method 2: Using the defining linear equation: m = 6 and points (1, –4) and (n, 170) y – y1 = m(x – x1) 170 – (–4) = 6(n – 1) 174 = 6(n – 1) 29 = n – 1 30 = n Therefore, the 30th term is 170.


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5. Find the sum of the first 125 terms defined by t1 = 4, tn = tn – 1 + 7. (3 marks) (Lesson 2) Answer: The arithmetic sequence is 4, 11, 18, . . . The common difference = 7 To find t125, use the formula tn = t1 + (n – 1)d Or t125 = 4 + (125 – 1)7 t125 = 4 + 875 – 7 t125 = 872

Sum of the series:

S125

4 8722

125 54 750=+

( )=

6. Find the total distance travelled by a ball in coming to rest, if it is dropped from a height

of 12 m and it rebounds 23

of its previous height every time it hits the ground (4 marks)

(Lesson 5) Answer:

down series:

12

23

1223

122

+ ( )+

( ) + . . .

down distance: 12

1 23

36−

= m

up series:

23

1223

1223

122 3

( ) +

( ) +

( ) + . . .

up distance = down distance – 12 = 24 m Total distance travelled = up distance + down distance = 24 m + 36 m = 60 m.

Sn

t n d

S

S

n= + −( )

= ( ) + −( )( )

= +

22 1

1252

2 4 125 1 7

1252

8

1

125

125 1124 7

54 750125

( )( )

=S


7. Consider the geometric sequence –1, 2, –4, . . . (Lesson 3)a) Write the function for the geometric sequence. (2 marks) Answer:

t

r

t t r

t

nn

nn

1

11

1

1

21

2

1 2

=−

=−

=−

=

=− −( )

−

−

b) Using this function, find t14 of the sequence. (1 mark) Answer: t14 = –1(–2)14–1

t14 = –1(–2)13

t14 = –1(–8192) t14 = 8192

c) Which term of the geometric sequence is the number –4096? (2 marks) Answer: tn = –1(–2)n–1

–4096 = –1(–2)n–1

4096 = (–2)n–1

(–2)12 = (–2)n–1

12 = n – 1 13 = n Therefore, –4096 is the 13th term in this geometric sequence.


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8. If possible, find the value of

64341

1

=

∞−

∑k

k

. (2 marks) (Lesson 5)

Answer:

r r r

t

Str

r

S

= <

=

=−

<∞

34

1

64

11

1

1

and satisfies the condition:

if

| |

| |

∞∞=−

=64

1 34

256


Module 2: Factoring and Rational Expressions (23 marks)

1. Factor the following expressions completely. (Lessons 1 and 2)a) x2y – 7xy + 10y (2 marks) Answer: = y(x2 – 7x + 10) = y(x – 2)(x – 5)

b) 4x2 + 12x + 9 (2 marks) Answer: = 4x2 + 6x + 6x + 9 = 2x(2x + 3) + 3(2x + 3) = (2x + 3)(2x + 3) = (2x + 3)2

c) 36x2 – 64y2 (1 mark) Answer: = 4(9x2 – 16y2) = 4(3x – 4y)(3x + 4y)

d) 25(x – 4)2 – 49(y – 5)2 (3 marks) Answer: 25(x – 4)2 – 49(y – 5)2

Let w = x – 4 t = y – 5 25w2 – 49t2

(5w – 7t)(5w + 7t) [5(x – 4) – 7(y – 5)][5(x – 4) + 7(y – 5)] [5x – 20 – 7y + 35][5x – 20 + 7y – 35] (5x – 7y + 15)(5x + 7y – 55)


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2. Without factoring, determine whether the binomial (x – 1) is a factor of the following polynomial. Explain. (2 marks) (Lesson 2) 2x2 – x – 1

Answer: If (x – 1) were a factor of this polynomial, then when x = 1 is substituted into the

polynomial, the polynomial would equal zero. 2(1)2 – 1 – 1 = 2 – 1 – 1 = 1 – 1 = 0

This polynomial equals zero when x = 1 is substituted in for x. Therefore, (x – 1) is a factor of this polynomial.

3. Create an equivalent rational expression for the following rational expression. Explain how you know the rational expression you created is equivalent to the original rational expression. State the non-permissible values of your rational expression. Do not simplify your answer. (2 marks) (Lesson 3)

xx 2

2

Answer: Answers may vary. A sample answer is:

xx

x xx x

+=

+( ) −( )−( )

2 2 222 2

The numerator and denominator are both multiplied by the same non-zero polynomial. Because any polynomial divided by itself is one, you are essentially multiplying the expression by 1. Anything multiplied by 1 is equal to itself.

The non-permissible values are x = 0 and x = 2.


4. Perform the indicated operation and simplify your answer. State the non-permissible values.

a)x

x xx

x x+

+ −+

− −

16 22

2

2 (4 marks) (Lesson 5)

Answer:

xx x

xx x

xx x

xx x

x x

+

+ −+

− −

=+

−( ) +( )+

−( ) +( )

= −( ) +(

16 2

12 3 2 1

2 3

2

2

2

2

LCD )) +( )

=+( ) +( )

−( ) +( ) +( )+

+( )−( ) +( ) +( )

=

x

x xx x x

x xx x x

x

1

1 12 3 1

32 1 3

2

22 3 2

3 2

2 1 32 3 1

4 2 12 3 1

+ + + +

−( ) +( ) +( )

=+ + +

−( ) +( ) +( )

x x xx x x

x x xx x x

Non-permissible values: x = 2, –3, –1

b)x xx x

x xx x

2

2

2

226

122 2 40

+ −

− −×

+ −

− − (3 marks) (Lesson 4)

Answer:

x xx x

x xx x

x xx x

x x

2

2

2

226

122 2 40

2 13 2

4

+ −

− −×

+ −

− −

=+( ) −( )−( ) +( )

×+( ) −33

2 20

2 13 2

4 32 5 4

2

( )− −( )

=+( ) −( )−( ) +( )

×+( ) −( )−( ) +( )

=

x x

x xx x

x xx x

xxx−

−( )1

2 5

Non-permissible values: x = 3, –2, 5, –4


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c)2 8

42 8

6

2

2

2

3 2 2x

xx xx x

xx x

−÷

− −

+×

− − (3 marks) (Lesson 4)

Answer:

2 84

2 86

2 44 2

8

2

2

2

3 2 2

2

2

3 2

2 2

xx

x xx x

xx x

xx

x xx x

xx

−÷

− −

+×

− −

=−( )

×+

− −×

−xx

x xx

x xx x

xx x

xx

−

=−( ) +( )

×+( )

−( ) +( )×

−( ) +( )

=−

6

2 2 24

12 1

83 2

43

2

2

Non-permissible values: x = 0, –1, 2, 3, –2

5. Create a rational expression with the following non-permissible values. Do not simplify your answer. (1 mark) (Lesson 3) –3 and –5

Answer: If –3 and –5 are non-permissible values, then (x + 3) and (x + 5) are factors of the

denominator of the rational expression. Answers may vary. However, (x + 3) and (x + 5) have to be factors of the denominator. A sample answer is:

13 5x x+( ) +( )


Module 3: Quadratic Functions (25 marks)

1. Given the graph below, answer the following questions. (Lessons 1 to 4)

y

x

2

2

2 2

a) What are the coordinates of the vertex? (1 mark) Answer: (1, –5)

b) What is the range? (1 mark) Answer: R: {y|y ³ –5} or [–5, ¥) or {y|y ³ –5, y Î Â}

c) What is the equation of the axis of symmetry? (1 mark) Answer: x = 1

d) State whether the graph has a maximum or a minimum value and what that value is. (1 mark)

Answer: It has a minimum value when y = –5.

e) Write a quadratic function in vertex form to represent this graph. (2 marks) Answer: Since the graph is normal width, the value of a is 1. y = (x – 1)2 – 5


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2. Match each equation to its corresponding graph. Place (a), (b), (c), and (d) next to its corresponding graph below. (1 mark each, for a total of 4 marks) (Lessons 1 to 4)a) y = 2(x – 2)2 – 2

b) y x=− −( ) +

12

2 22

c) y x=− +( ) +

12

2 22

d) y = 2(x + 2)2 – 2

______

x

y

0

2

2

_____

x

y

0

2

2

______

x

y

0 2

2

______

x

y

0

2

2

3. For what value of k is the expression x2 – 11x + k a perfect square trinomial? (1 mark) (Lesson 5)

Answer:

−

=

112

1214

2

(b) (c)

(d)(a)


4. Consider the parabola y = 3x2 + 12x – 36. (Lesson 5)a) Write the function in vertex form by completing the square. (2 marks) Answer: y = 3(x2 + 4x) – 36 y = 3(x2 + 4x + 4) – 36 – 4(3) y = 3(x + 2)2 – 48

b) Find the coordinates of the vertex. (1 mark) Answer: (–2, –48)

c) Find the y-intercept. (1 mark) Answer: Let x = 0 y = 3x2 + 12x – 36 y = 3(0)2 + 12(0) – 36 y = –36

d) State the domain. (1 mark) Answer: D: {x|x Î Â} or (–¥, ¥)

e) Write an equation for the axis of symmetry. (1 mark) Answer: x = –2


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f) Sketch the graph. (2 marks) Answer: The vertex is (–2, –48). The curve opens up and has a narrow width. The y-values are

three times the normal ones.

y

x

10

22

10

20

30

40

50

46

g) Find the x-intercepts from the graph. (1 mark) Answer: The x-intercepts are located at x = –6 and x = 2.


5. A projectile is shot straight up from a height of 40 m with an initial velocity of 30 m/s. Its height in metres after t seconds is given by h(t) = 40 + 30t – 5t2. (1.5 marks each, for a total of 3 marks) (Lesson 7)a) After how many seconds does the projectile reach its maximum height? Answer: h(t) = –5t2 + 30t + 40

Time at maximum height =

− =−

−( )=

ba2

302 5

3 seconds

Note: This question could also be solved by completing the square to find the vertex at (3, 85), where 3 seconds is the answer to (a) and 85 metres is the answer to (b).

b) Find the maximum height above the ground that the ball reaches. Answer: Maximum height = h(3) = –5(3)2 + 30(3) + 40 = 85 metres

6. Consider the following quadratic functions. Determine how many x-intercepts the corresponding graph has by considering the values of a and q. (1 mark each, for a total of 2 marks) (Lesson 6)a) y = –2(x + 3)2 – 15 Answer: In this equation, a = –2 < 0, and q = -15 < 0. Therefore, as both a and q are less than

zero, the corresponding graph has no x-intercepts. Or The graph opens down and the vertex is below the x-axis. There are no x-intercepts.

b) y = 9(x + 2)2

Answer: In this equation, a = 9 > 0, and q = 0. Therefore, the corresponding graph has one (1)

x-intercept, as the vertex is on the x-axis. Or The graph opens up and the vertex is on the x-axis. There is one x-intercept.


Name:

Module 4: Solving Rational and Quadratic Equations (28 marks)

1. Find the roots of the following equations. Explain which method you used to solve each equation and why. (Lessons 1 to 4)a) x2 – 9x + 18 = 0 (2 marks) Answer: This equation is easily factorable. (x – 6)(x – 3) = 0 x = 6, 3

b) x2 – 5x – 1 = 0 (3 marks) Answer: This equation is not easily factorable. Also, if you were to complete the square, there

would be fractions. Therefore, the quadratic formula should be used. Note: If you use the completing-the-square method, you should still arrive at the

correct answer.

xb b ac

a

x x

a b c

x

x

=− ± −

− − =

= =− =−

=± −( ) − ( ) −( )

( )

2

2

2

42

5 1 0

1 5 1

5 5 4 1 12 1

, ,

==± +

=±

=+ −

5 25 42

5 292

5 292

5 292

x

x ,

Note: Unless specifically mentioned, you may use any method to solve quadratic equations.


2. Use the quadratic formula to find the roots of 2x2 + 6x – 1 = 0. (2 marks) (Lesson 4) Answer:

xb b ac

aa b c

x

x

x

=− ± −

= = =−

=− ± − ( ) −( )

( )

=− ± +

=−

2

2

42

2 6 1

6 6 4 2 12 2

6 36 84

, ,

66 444

6 4 114

6 2 114

3 112

3 112

3 112

±

=− ±

=− ±

=− ±

=− + − −

x

x

x

x ,


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3. Solve x2 – x – 3 = 0 by completing the square. (3 marks) (Lesson 4) Answer:

x x

x x

x x

x

x

x

2

2

2

2

3 0

3

14

314

12

134

12

134

12

132

− − =

− =

− + = +

−

=

− =±

= ±

xx=+ −1 13

21 13

2,

4. Solve (x – 7)2 = 36 by taking square roots. (2 marks) (Lesson 2) Answer: (x – 7)2 = 36 x – 7 = ±6 x = 7 ± 6 x = 7 + 6, 7 – 6 x = 13, 1


5. Solve (x – 2)2 – 1 = 0 by graphing. (2 marks) (Lesson 1) Answer: Let y = (x – 2)2 – 1 Vertex is (2, –1); the curve opens up and its width is normal.

y

x4

2

2

4

The x-intercepts (or zeros, or roots) of this quadratic equation are 3 and 1.


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6. Solve for x. State the non-permissible values of the rational equation. (4 marks) (Lesson 6)

xx

xx

−+

− =+( )−

23

34 3

2

Answer:

xx

xx

x x

x

−+

− =+( )−

= +( ) −( )

=

23

34 3

2

3 2LCD

Non-permissible values are −−

−+

+( ) −( )− +( ) −( )=+( )−

+( ) −( )

−(

3 2

23

3 2 3 3 24 3

23 2

2

, .

xx

x x x xxx

x x

x )) −( )− − + −( )= +( ) +( )

− +( )− + −( )= +

x x x x x x

x x x x x

2 3 2 3 6 4 3 3

4 4 3 6 4 6

2

2 2 2 xx

x x x x x x

x x x x

+( )

− + − − + = + +

− − + = + +

=

9

4 4 3 3 18 4 24 36

2 7 22 4 24 36

0 6

2 2 2

2 2

xx x

x x x

x x x

x x

2

2

31 14

0 6 3 28 14

0 3 2 1 14 2 1

0 3 14 2 1

+ +

= + + +

= +( )+ +( )

= +( ) +( ))

=− −

x143

12

,


7. The discriminant of the quadratic equation 4x2 + 2x – k = 0 is 84. (1 mark each, for a total of 2 marks) (Lesson 5)a) Find the value of k. Answer: a = 4, b = 2, c = –k The discriminant = b2 – 4ac 84 = (2)2 – 4(4)(–k) 84 = 4 + 16k 80 = 16k 5 = k

b) State the nature of the roots without solving for the roots. Answer: As the discriminant is positive and not a perfect square, there are 2 real, irrational,

unequal roots.


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8. Tia can ride her bicycle 1 km/h faster than Denzel. They begin at the same spot and at the same time, travelling in opposite directions. After travelling for the same length of time, they stop and realize that Tia has travelled 21 km and Denzel has travelled 19 km. How fast was Denzel travelling? (4 marks) (Lesson 6)

Recall: Distance = Rate ´ Time, or Time =

DistanceRate

Answer: Let r = Denzel’s rate.

Rate ´ Time = Distance

Tia r + 121

1r 21

Denzel r 19r

19

Tia’s time of travel = Denzel’s time of travel

211

19

1

211

119

1

21 19 1

2

r r

r r

rr r

rr r

r r

+=

= +( )( )

++( )( )= +( )( )

= +( )

LCD

11 19 19

2 19

9 5

r r

r

r

= +

=

= .

Therefore, Denzel was travelling at 9.5 km/h.


9. Andrea can mow a lawn by herself in 4 hours. Samantha can mow the same lawn in 3 hours. If the girls work together, how long will it take to mow this lawn? (4 marks) (Lesson 6)

Answer: The following equation can be used to solve this problem: (per hour rate) ´ (hours worked) = portion of lawn mowed

In one hour, the Andrea can mow

14

of the lawn. Therefore, Andrea has a per-hour rate

of 14

.

In one hour, Samantha can mow

13

of the lawn. Therefore, Samantha has a per-hour rate

of 13

.

Per Hour Rate

Hours Worked

Portion of Lawn Mowed

Andrea 14

x x4

Samantha 13

x x3

When both girls are working together, the portion of the lawn mowed in x hours is the whole lawn, and is represented in the equation as 1. The equation becomes:

x x

x x

x

x

x

4 31

12

312

412

1

712

1

7 12

127

1 7

+ =

=

+ =

=

=

= ≈

LCD

hours.

Therefore, both of these girls working together can mow the lawn in approximately 1.7 hours.

G r a d e 11 P r e -C a l C u l u s M a t h e M at i C s (30s)6 of 26 Grade 11 Pre-Calculus Mathematics 3. Write the defining linear function of the following arithmetic sequence.

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